I think the airplan flies in same speed 400
At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is
H = 50 - (1/2 G Q²) ,
and the height of the stone that was tossed straight up
from the ground is
H = 20Q - (1/2 G Q²) .
The stones meet when them's heights are equal,
so that's the time when
<span>50 - (1/2 G Q²) = 20Q - (1/2 G Q²) .
This is looking like it's going to be easy.
Add </span><span>(1/2 G Q²) to each side.
Then it says
50 = 20Q
Divide each side by 20: 2.5 = Q .
And there we are. The stones pass each other
2.5 seconds
after they are simultaneously launched.
</span>
Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for 
DAC:
⇒ 
Now for the BAC:
⇒ 
Now, differentiating w.r.t x:
![\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%20%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Btan%5E%7B-1%7D%20%5Cfrac%7Bd%20%2B%20h%7D%7Bx%7D%20-%20%20tan%5E%7B-1%7D%20%5Cfrac%7Bd%7D%7Bx%7D%5D)
For maximum angle, 
= 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 = 
After solving the above eqn, we get
x = 
The observer should stand at a distance equal to x =
Explanation:
This is true because at maximum height, the velocity is 0
Answer:
Explanation:
F = kQq/r²
r = √(kQq/F)
a) r = √(8.899(10⁹)(8)(4) / 18(10¹³)) = 0.0397749... m
r = 40 mm
b) r = √(8.899(10⁹)(12)(3) / 18(10¹³)) = 0.0421876... m
r = 42 mm
